So many graphs!

Algebra Level 4

Consider a quadratic expression in $x$ having real roots $\alpha_{1}$ , $\alpha_{2}$ such that they both are within the interval $(-(\alpha_{1}+\alpha_{2}),(\alpha_{1}+\alpha_{2}))$ , then enter the numbers of the graphs which are plausible for the expression on $y$ -axis and $x$ on the $x$ -axis.

Concatenate the numbers in ascending order. For example, if graphs 1,3 and 2 are plausible, enter 123.

$\large 1.$ $\large 2.$ $\large 3.$ $\large 4.$ $\large 5.$

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The answer is 235.

2 solutions

Rohith M.Athreya
Jan 14, 2017

the said interval is $(-(\alpha_{1}+\alpha_{2}),(\alpha_{1}+\alpha_{2})$

this implies $ab<0$ if the quadratic expression is $f(x)=ax^2+bx+c$ .

using the condition on the sign of $f((\alpha_{1}+\alpha_{2}))$ , we get $c>0$ if $a>0$ and c is negative otherwise

now that product and sum of roots are positive, the roots have to be positive

why is ab<0 ?,both roots can be negative

Anirudh Sreekumar - 4 years, 4 months ago

negative sum of roots is lesser than positive counterpart thats why

Rohith M.Athreya - 4 years, 4 months ago

but what if reversed intervals are allowed,if $(b,a)$ is an inteval with $b>a$ ,both roots still lie in it

Anirudh Sreekumar - 4 years, 4 months ago

@Anirudh Sreekumar – it is convention that u specify (2,3)

if u unconventionally use (3,2), there are a few drawbacks in using your system

however,since that doesnt impact the answer, it is matter of convention

Rohith M.Athreya - 4 years, 4 months ago

Kushal Bose
Jan 14, 2017

The roots should have same sign. either both neagative or both positive.

Let's see his with an example .Consider , $a_1=-1$ and $a_2=5$ .The given range will be $-4,4$ which is not correct

So, in the grapg $2,3$ are suitible options.

The graph has asingle distinct which also satisfies.

So the final answer is $235$

the roots are always positive can be proved by rigour

Rohith M.Athreya - 4 years, 5 months ago